Question The matrix A=[a h; h b] is transformed to the diagonal form D = T −1AT, where T=[cos θ sin θ; − sin θ cos θ]. Find value of θ which gives this diagonal transformation.

A=\left[\begin{array}{ll}a & h \\h & b\end{array}\right]\text { Diagaral form of } A=\left[\begin{array}{ll}a & 0 \\0 & b\end{array}\right]D=\left[\begin{array}{ll}a & 0 \\0 & b\end{array}\right]=T^{-1} A TGiven T=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]\begin{array}{l}\text { A AT }=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\\sin \theta & \cos \theta\end{array}\right]\left[\begin{array}{ll}a \cos \theta+h \sin \theta & -a \sin \theta+h \cos \theta \\h \operatorname{cs} \theta+b \sin \theta & -h \sin \theta+b \cos \theta\end{array}\right] \\=\left[\begin{array}{lr}a \operatorname{cs}^{2} \theta+h \cos \theta \sin \theta-h c \cos \theta \sin \theta-b \sin ^{2} \theta & -a \sin \theta \cos \theta+h \cos ^{2} \theta+h \sin ^{2} \theta \\a \sin \theta \cos \theta+h \sin ^{2} \theta+h \cos ^{2} \theta+b \cos \theta \sin \theta & -a \sin ^{2} \theta+h \sin \theta \cos \theta \\& -h \cos \theta \sin \theta+b \cos ^{2} \theta\end{array}\right] \\=\left[\begin{array}{ll}a \cos ^{2} \theta-b \sin ^{2} \theta & -a \sin \theta \cos \theta+h\left(\cos ^{2} \theta+\sin ^{2} \theta\right)-b \sin \theta \operatorname{cs} \theta \\a \sin \theta \cos \theta[a+b]+h\left[\cos ^{2} \theta+\sin ^{2} \theta\right] & -a \sin ^{2} \theta+b \cos ^{2} \theta\end{array}\right] \\=\left[\begin{array}{cc}a \operatorname{cs}^{2} \theta-b \sin ^{2} \theta & -a \sin \theta \cos \theta+h-b \sin \theta \cos \theta . \\\sin \theta \operatorname{cs} \theta(a+b)+h & -a \sin ^{2} \theta+b \operatorname{cs}^{2} \theta\end{array}\right] . \\\end{array}To Traustiom Thir to a Diggnal matrine=\left[\begin{array}{cc}a \cos ^{2} \theta-b \sin ^{2} \theta & \sin \theta c \operatorname{cs} \theta(-a-b)+h \\\sin \theta c \operatorname{cs} \theta(a+b)+h & -a \sin ^{2} \theta+b \cos ^{2} \theta\end{array}\right]escocos (-x-b) x=h\begin{array}{l}=\left[\begin{array}{cc}a \csc ^{2} \theta-b \sin ^{2} \theta & 0 \\0 & -a \sin ^{2} \theta+b c^{2} \theta\end{array}\right] \\\theta=90^{\circ} \text { (or) } 0^{\circ}, h=0 \\=\left[\begin{array}{cc}a-0 & 0 \\0 & -a\end{array}\right] \text { when } \theta=0^{\circ} \\=\left[\begin{array}{cc}-b & 0 \\0 & +b\end{array}\right] \text { shen } \theta=90^{\circ}\end{array} ...